# Element has Idempotent Power in Finite Semigroup

## Theorem

Let $\struct {S, \circ}$ be a finite semigroup.

For every element in $\struct {S, \circ}$, there is a power of that element which is idempotent.

That is:

$\forall x \in S: \exists i \in \N: x^i = x^i \circ x^i$

## Proof

From Finite Semigroup Equal Elements for Different Powers, we have:

$\forall x \in S: \exists m, n \in \N: m \ne n: x^m = x^n$

Let $m > n$.

Let $n = k, m = k + l$.

Then:

$\forall x \in S: \exists k, l \in \N: x^k = x^{k + l}$

Now we show that:

$x^k = x^{k + l} \implies x^{k l} = x^{k l} \circ x^{k l}$

That is, that $x^{k l}$ is idempotent.

First:

 $\displaystyle x^k$ $=$ $\displaystyle x^{k + l}$ $\displaystyle \leadsto \ \$ $\displaystyle x^k \circ x^l$ $=$ $\displaystyle x^{k + l} \circ x^l$ both sides ${} \circ x^l$ $\displaystyle \leadsto \ \$ $\displaystyle x^{k + l}$ $=$ $\displaystyle x^{k + 2 l}$ Index Laws for Semigroup: Sum of Indices‎ $\displaystyle \leadsto \ \$ $\displaystyle x^k$ $=$ $\displaystyle x^{k + 2 l}$ as $x^k = x^{k + l}$

From here we can easily prove by induction that:

$\forall n \in \N: x^k = x^{k + n l}$

In particular:

$x^k = x^{k + k l} = x^{k \paren {l + 1} }$

There are two cases to consider:

$(1): \quad$ If $l = 1$, then $x^k = x^{k \paren {l + 1} } = x^{2 k} = x^k \circ x^k$, and $x^{k l} = x^k$ is idempotent.
$(2): \quad$ If $l > 1$, then:
 $\displaystyle x^k$ $=$ $\displaystyle x^{k + k l}$ $\displaystyle$ $=$ $\displaystyle x^{k \paren {l + 1} }$ $\displaystyle \leadsto \ \$ $\displaystyle x^k \circ x^{k \paren {l - 1} }$ $=$ $\displaystyle x^{k \paren {l + 1} } \circ x^{k \paren {l - 1} }$ both sides ${} \circ x^{k \paren {l - 1} }$ $\displaystyle \leadsto \ \$ $\displaystyle x^{k + k l - k}$ $=$ $\displaystyle x^{kl + k + k l - k}$ Index Laws for Semigroup: Sum of Indices $\displaystyle \leadsto \ \$ $\displaystyle x^{k l}$ $=$ $\displaystyle x^{k l + k l}$ $\displaystyle$ $=$ $\displaystyle x^{k l} \circ x^{k l}$

and again, $x^{k l}$ is idempotent.

$\blacksquare$